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How does one write out the full mathematical expression for an ARIMA(3,0,2) - GARCH(1,1) model if the following list represents the estimates of the coefficients?

Coefficient estimates:

mu: 0.1, ar1: 0.2, ar2: 0.3, ar3: 0.4, ma1: 0.5, ma2: 0.6, omega: 0.7, alpha1: 0.8, beta1: 0.9

Richard Hardy
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berkshire
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  • See [What is the difference between GARCH and ARMA?](https://stats.stackexchange.com/questions/41509/what-is-the-difference-between-garch-and-arma). – Richard Hardy Apr 09 '18 at 05:21

1 Answers1

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If I understood correctly you asked about the formula for an ARIMA and a GARCH process based on those coefficients. Clearly there is no unique way to assign labels to parameters, but these are two common specifications:

ARIMA(3,0,2):

$ Y_t = \mu + a_1 Y_{t-1} + a_2 Y_{t-2} + a_3 Y_{t-3} + \epsilon_t + m_1 \epsilon_{t-1} + m_2 \epsilon_{t-2} $

GARCH(1,1):

$ \sigma^2 = \omega + \alpha_1 r^2_{t-1} + \beta_1 \sigma^2_{t-1} $

where $r_t = \sqrt{\sigma^2_t}Z_t$, with $Z_t \sim F(mean=0, var=1)$.

Ale
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  • Either this or $Y_t = \mu + a_1 Y_{t-1} + a_2 Y_{t-2} + a_3 Y_{t-3} + \epsilon_t + m_1 \epsilon_{t-1} + m_2 \epsilon_{t-2}$ replaced by $(Y_t-\mu) = a_1 (Y_{t-1}-\mu) + a_2 (Y_{t-2}-\mu) + a_3 (Y_{t-3}-\mu) + \epsilon_t + m_1 \epsilon_{t-1} + m_2 \epsilon_{t-2}$. Base R uses the second expression for pure ARMA models, but `rugarch` might be using the first one for ARMA-GARCH models (I am not sure). – Richard Hardy Apr 09 '18 at 16:44